Pulley From The Other World

Really This is an Interesting question ! @Ishan Singh create very good question Finally I'm able to solved This question after a wrong attempt .

See Following Images first carefully !

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Since small rope element which we consider is in equilibrium ( Moving with constant velocity ) So we Balanced Forces in X and Y direction Respectively.

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X-axis :

$(T\quad +\quad dT)\cos { (\cfrac { d\theta }{ 2 } ) } \quad \quad =\quad \quad T\cos { (\cfrac { d\theta }{ 2 } ) } \quad +\quad { df }_{ net }\quad \quad \\ \\ \because \quad \cos { (\cfrac { d\theta }{ 2 } ) } \quad \approx \quad 1\\ \\ \Rightarrow \quad dT\quad =\quad { df }_{ net }\quad \quad \quad .\quad .\quad .\quad (1)$ .

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Y-axis :

$T\sin { \cfrac { d\theta }{ 2 } } \quad +\quad (T\quad +\quad dT)\sin { \cfrac { d\theta }{ 2 } } \quad =\quad N\\ \\ \because \quad \quad \sin { \cfrac { d\theta }{ 2 } } \approx \quad \cfrac { d\theta }{ 2 } \\ \\ \Rightarrow \quad Td\theta \quad +\quad dT(\cfrac { d\theta }{ 2 } )\quad =\quad N\\ \\ \because \quad dT(\cfrac { d\theta }{ 2 } )\quad \approx \quad 0\\ \\ \Rightarrow \quad Td\theta \quad =\quad N\quad \quad .\quad .\quad .\quad .\quad (2)$ .

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Also

${ df }_{ net }\quad =\quad \mu N\quad \quad .\quad .\quad .\quad .\quad (3)$ .

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using all 3 equations we get relation :

$\cfrac { dT }{ T } \quad =\quad \mu d\theta \\ \\ \int _{ mg }^{ F }{ \quad \cfrac { dT }{ T } } \quad =\quad \mu \int _{ 0 }^{ \pi }{ d\theta } \\ \\ F\quad =\quad mg{ \times e }^{ \pi \mu }\quad =\quad 10709.3\quad \\ \\ Q.E.D$ .

Try to prove

X = $\displaystyle mg \times e^{\pi \mu}$

$\mu$ - frictinal coefficient

only thing bothering me is the angle, i think it would be slightly less than pi and we cannot neglect it since it varies exponentially .